# Sum six unique digits to make 20

You have 6 boxes. You can use the digits from 1 to 9 but not 0. Digit repetition is not allowed. The total sum of the numbers/digits should be 20.

$$\huge\square\square\square\\ \huge\square\square\square$$

It's not possible. The sum of the 6 smallest digits (1 through 6) is 21, already greater than 20.

• So, are we allowed to use negative numbers to solve this? – Mehmood Ali Nov 4 '15 at 17:29
• The question is yours, how am I supposed to know. Haha. – Fimpellizieri Nov 4 '15 at 18:01
• Yeah, it is my question but can we use negative numbers to solve this according to mathematics rules? Is that possible? – Mehmood Ali Nov 4 '15 at 18:04
• Digits are not negative or positive, I am not sure I understand you. – Fimpellizieri Nov 4 '15 at 18:05
• Alright, I got you. – Mehmood Ali Nov 4 '15 at 18:14

Seems pretty easy to me.

$$\huge[9][1][8]\\ \huge[2]\square\square$$

The rules don't say I need to fill in all of the boxes.

If we add the tag, I have a solution:

Write 4, 5, and 8 in the first three boxes
Cut off three of your fingers (also called "digits") and place one in each of the other three boxes.
Now you have 6 digits (three numbers, three fingers) in the boxes
Each finger has a value of 1 so the sum is $1+1+1+4+5+8=20$

Alternatively, if fingers have a value of 0, then us the numbers 2, 3, 4, 5, 6 and only cut off one finger (value of 0). Now the sum is $0+2+3+4+5+6=20$

• That's pretty bloody. I'd rather skip the puzzle than lose three fingers. – Illyasviel Nov 5 '15 at 15:38

Inspired by @AlbertRenshaw's post:

Use hexadecimal. Then $2 + 4 + 5 + 6 + 7 + 8 = 20_{16}\;(=32_{10})$.

This can actually work with any base from 11 to 19:

$1 + 2 + 3 + 4 + 5 + 7 = 20_{11}\;(=22_{10})$
$1 + 2 + 3 + 4 + 5 + 9 = 20_{12}\;(=24_{10})$
$1 + 2 + 3 + 4 + 7 + 9 = 20_{13}\;(=26_{10})$
$1 + 2 + 3 + 6 + 7 + 9 = 20_{14}\;(=28_{10})$
$1 + 2 + 3 + 7 + 8 + 9 = 20_{15}\;(=30_{10})$
$1 + 4 + 5 + 7 + 8 + 9 = 20_{17}\;(=34_{10})$
$1 + 5 + 6 + 7 + 8 + 9 = 20_{18}\;(=36_{10})$
$3 + 5 + 6 + 7 + 8 + 9 = 20_{19}\;(=38_{10})$

# 1,2,3,4,5,7

And the puzzle is in base 9.

• Alas, if the puzzle were in base 9, 9 would not be a valid digit. – Ian MacDonald Nov 5 '15 at 15:03
• @IanMacDonald 9 = ∞ in base 9. xD – Albert Renshaw Nov 5 '15 at 16:54
• That's not quite how math works. :P – Ian MacDonald Nov 5 '15 at 17:05

## protected by Beastly GerbilMar 24 '17 at 18:11

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